2014-01-14 44 views
0

我有一个简单的PHP代码来显示与数据库的连接是否已建立。它有一个HTML部分,但它似乎没有建立连接。使用的数据库是MySql。这些表格不是作为程序的一部分创建的。无法使用Mysql在PHP中建立与数据库的连接

需要一些建议....

谢谢....提前。

这里是代码...

<h1>LOGIN PAGE </h1> 
<form action="" method="post"> 
Username <input name="usname" type="text" /> 
Password <input name="paswd" type="password" /> 
<input name="submit" type="submit" /> 
</form> 

<?php 
if (isset($_POST['submit'])) { 
    $uname = $_POST['usname']; 
    $pswd = $POST['paswd']; 
    $conn = mysqli_connect("localhost","root","archana","details"); 
    if(mysqli_connect_errorno()) 
    { 
     echo "Failed to connect to Mysql" ; 
    } 
    else 
    { 
     echo "connection established"; 
    } 
    $data = mysqli_query($conn, "SELECT * FROM user"); 
    while($a = mysqli_fetch_array($data)) 
    { 
     echo $a['username']; 
    } 
    mysqli_close($conn); 
} 
?> 
+1

确保你的用户和密码是正确的mysqli_connect。同时检查细节数据库是否存在。 –

+0

当前代码回显是什么?它会显示这条消息吗?'“连接到Mysql失败”' –

+3

'mysqli_connect_errorno()'应该是'mysqli_connect_errno()' –

回答

0

我重写基于马塞尔巴尔泽尔代码注释

<h1>LOGIN PAGE </h1> 
<form action="" method="post"> 
Username <input name="usname" type="text" /> 
Password <input name="paswd" type="password" /> 
<input name="submit" type="submit" /> 
</form> 

<?php 
if (isset($_POST['submit'])) { 
    $uname = $_POST['usname']; 
    $pswd = $POST['paswd']; 
    $conn = mysqli_connect("localhost","root","archana","details"); 
    if(mysqli_connect_errno()) 
    { 
     echo "Failed to connect to Mysql. Error: ".mysqli_connect_errno() ; 
    } 
    else 
    { 
     echo "connection established"; 
    } 
    $data = mysqli_query($conn, "SELECT * FROM user"); 
    while($a = mysqli_fetch_array($data)) 
    { 
     echo $a['username']; 
    } 
    mysqli_close($conn); 
} 
?> 
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